pub fn new_birthday_probability(n: u32) -> f64 {
    // TODO: 这里写逻辑
    if n < 2 {
        return 0.0;
    }

    let days_in_year = 365;
    let mut no_shared_birthday_probability = 1.0;

    // 计算没有重复生日的概率
    for i in 0..n {
        no_shared_birthday_probability *= (days_in_year - i) as f64 / days_in_year as f64;
    }

    // 至少有两人生日相同的概率
    let shared_birthday_probability = 1.0 - no_shared_birthday_probability;

    // 保留四位小数
    (shared_birthday_probability * 10000.0).round() / 10000.0
}
